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Proof step 12 from proof bitsxor-zeros (previous step)

bits0 = bits0.first ', 1 ≤ # bits, ∀ bits, bits1. # bits1 < 1 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ bits = [], bits0 = [], bitsxor(bits, bits0) = bits, true ∈ bits0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
bits0 = boolvar + bits2, bits0 = boolvar ', 1 ≤ # bits, ∀ bits, bits1. # bits1 < 1 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ bits = [], bits0 = [], bitsxor(bits, bits0) = bits, true ∈ bits0

Rule argument

seq: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0) subst: [bits, boolvar, bits0] → [bits0, boolvar, bits2] rotate: T spec: list-data inst: bitlista name: elim-1

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0); (elim) Used terms: bits0, boolvar, bits2